Two capacitors $C_1 = 2 \mu F$ and $C_2 = 4 \mu F$ are connected in a circuit as shown in the figure. The potential difference $(V_A - V_B)$ is....$V$

The two plates $X$ and $Y$ of a parallel plate capacitor of capacitance $C$ are given a charge of amount $Q$ each. $X$ is now joined to the positive terminal and $Y$ to the negative terminal of a cell of $emf$ $E = Q/C$.

$A$ metal ball of radius $R$ is placed concentrically inside a hollow metal sphere of inner radius $2R$ and outer radius $3R$. The ball is given a charge $+2Q$ and the hollow sphere is given a total charge $-Q$. The electrostatic potential energy of this system is:

$n$ identical capacitors are connected in parallel and charged to a potential $V$. Now,they are disconnected and reconnected in series. What will be the total energy and potential difference of the combination?

Charge on the capacitor in the given circuit in steady state condition is :- ............... $\mu C$

$A$ parallel plate capacitor of plate area $A$ and plate separation $d$ is charged to a potential $V$ and then the battery is disconnected. $A$ slab of dielectric constant $k$ is then inserted between the plates of the capacitor so as to fill the space between the plates. If $Q$,$E$,and $W$ denote respectively the magnitude of charge on each plate,the electric field between the plates (after the slab is inserted),and the work done on the system in the process of inserting the slab,then state the incorrect relation from the following: